# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
152940 | Ruxandra985 | popa (BOI18_popa) | C++14 | 10 ms | 380 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <cstdio>
#include "popa.h"
/// daca iese asa sunt zeita si intuitia mea este aur:)
int solve (int n , int *l , int *r){ /// return root
int i,lant,root;
int bgn[1010] , ed[1010] , tt[1010] , fii[1010];
for (i=1;i<=n;i++){
tt[i] = l[i] = r[i] = -1; /// reset all
}
/// separam in lanturi
lant = 1;
bgn[1] = ed[1] = 1;
for (i=2;i<=n;i++){ /// n - 1
if (query(bgn[lant] , i , i , i)){ /// i apartine lantului curent
ed[lant]++;
tt[i-1] = i;
}
else { /// incepem alt lant
lant++;
bgn[lant] = ed[lant] = i;
}
}
/// mai bine obtin l si r din tt
/// cum reusesc eu sa conectez aceste lanturi
/// uniunile dintre lanturi se fac clar la nivelul lui ed
/// pentru ca au indici consecutivi in s
root = ed[1];
for (i=2;i<=lant;i++){ /// ce fac cu lantul asta
/// daca pot sa il unesc pe ed[i-1] cu ed[i]
if (query(ed[i-1] , ed[i] , ed[i-1],ed[i-1])){
tt[ed[i]] = ed[i-1]; /// unesti intervalele astea
}
else { /// daca nu pot cu aia , il fac pe ed[i] radacina
tt[root] = ed[i];
root = ed[i];
ed[i]--;
}
}
for (i=1;i<=n;i++){
if (tt[i]!=-1){
if (l[tt[i]]==-1)
l[tt[i]] = i;
else r[tt[i]] = i;
}
}
return root;
}
Compilation message (stderr)
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |